Question: $ABCD$ is a trapezoid with the measure of base $\overline{AB}$ twice the measure of the base $\overline{CD}$. Point $E$ is the point of intersection of the diagonals. The measure of diagonal $\overline{AC}$ is 11. Find the length of segment $\overline{EC}$. Express your answer as a common fraction.

[asy]

size(200);
pair p1,p2,p3,p4;
p1 = (0,0); p2 = (2.5, 4); p3 = (7.5,4); p4 = (10,0);

draw(p1--p2--p3--p4--cycle);
draw(p1--p3); draw(p2--p4);
label("$A$", p1, SW);
label("$D$", p2, NW);
label("$C$", p3, NE);
label("$B$", p4, SE);
label("$E$", (5,2.5) , S);

[/asy]
Solution: Since the bases of the trapezoid are $\overline{AB}$ and $\overline{CD}$, these two line segments must be parallel. Now, since $\overline{AC}$ intersects these two parallel lines, $\angle DCE$ and $\angle BAE$ are alternate interior angles and therefore must be congruent. Similarly, $\overline{DB}$ intersects the bases, so $\angle CDE$ and $\angle ABE$ are congruent. We have two pairs of congruent angles, so $\triangle DCE \sim \triangle BAE$ by the Angle-Angle Similarity Theorem.

Sides of similar triangles are proportional, so since the lengths of sides $\overline{AB}$ and $\overline{CD}$ are related in a $2:1$ proportion, we also have that $EA/EC=2/1$, so the length of $\overline{EC}$ must be $1/3$ that of $\overline{AC}$. Since $\overline{AC}$ has length $11$, $\overline{EC}$ must have length $\dfrac{1}{3} \cdot 11 = \boxed{\dfrac{11}{3}}$.